Question

(20x^3-7x^2+3x-7)/-13x^2-5

Answer 1

Answer:

x = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (30 sqrt(1462809) + 36253)^(1/3)) + 7/60 or x = 7/60 - 1/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) (131 (-1)^(1/3) + (30 sqrt(1462809) + 36253)^(2/3)) or x = 1/60 (131 (-1/(36253 + 30 sqrt(1462809)))^(1/3) + (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)) + 7/60

see atachement it's more legible.

Step-by-step explanation:

Solve for x:

(20 x^3 - 7 x^2 + 3 x - 7)/(-13 x^2 - 5) = 0

Hint: | Multiply both sides by a polynomial to clear fractions.

Multiply both sides by -13 x^2 - 5:

20 x^3 - 7 x^2 + 3 x - 7 = 0

Hint: | Look for a simple substitution that eliminates the quadratic term of 20 x^3 - 7 x^2 + 3 x - 7.

Eliminate the quadratic term by substituting y = x - 7/60:

-7 + 3 (y + 7/60) - 7 (y + 7/60)^2 + 20 (y + 7/60)^3 = 0

Hint: | Write the cubic polynomial on the left hand side in standard form.

Expand out terms of the left hand side:

20 y^3 + (131 y)/60 - 36253/5400 = 0

Hint: | Write the cubic equation in standard form.

Divide both sides by 20:

y^3 + (131 y)/1200 - 36253/108000 = 0

Hint: | Perform the substitution y = z + λ/z.

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-36253/108000 + (131 (z + λ/z))/1200 + (z + λ/z)^3 = 0

Hint: | Transform the rational equation into a polynomial equation.

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ + 131/1200) - (36253 z^3)/108000 + z^2 (3 λ^2 + (131 λ)/1200) + λ^3 = 0

Hint: | Find an appropriate value for λ in order to make the coefficients of z^2 and z^4 both zero.

Substitute λ = -131/3600 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - (36253 u)/108000 - 2248091/46656000000 = 0

Hint: | Solve for u.

Find the positive solution to the quadratic equation:

u = (36253 + 30 sqrt(1462809))/216000

Hint: | Perform back substitution on u = (36253 + 30 sqrt(1462809))/216000.

Substitute back for u = z^3:

z^3 = (36253 + 30 sqrt(1462809))/216000

Hint: | Take the cube root of both sides.

Taking cube roots gives 1/60 (36253 + 30 sqrt(1462809))^(1/3) times the third roots of unity:

z = 1/60 (36253 + 30 sqrt(1462809))^(1/3) or z = -1/60 (-36253 - 30 sqrt(1462809))^(1/3) or z = 1/60 (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)

Hint: | Perform back substitution with y = z - 131/(3600 z).

Substitute each value of z into y = z - 131/(3600 z):

y = 1/60 (30 sqrt(1462809) + 36253)^(1/3) - 131/(60 (30 sqrt(1462809) + 36253)^(1/3)) or y = -1/60 (-30 sqrt(1462809) - 36253)^(1/3) - (131 (-1)^(2/3))/(60 (30 sqrt(1462809) + 36253)^(1/3)) or y = 131/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) + 1/60 (-1)^(2/3) (30 sqrt(1462809) + 36253)^(1/3)

Hint: | Simplify each solution.

Bring each solution to a common denominator and simplify:

y = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (36253 + 30 sqrt(1462809))^(1/3)) or y = -1/60 (-1/(36253 + 30 sqrt(1462809)))^(1/3) ((30 sqrt(1462809) + 36253)^(2/3) + 131 (-1)^(1/3)) or y = 1/60 (131 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) + (-1)^(2/3) (30 sqrt(1462809) + 36253)^(1/3))

Hint: | Perform back substitution on the three roots.

Substitute back for x = y + 7/60:

Answer: x = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (30 sqrt(1462809) + 36253)^(1/3)) + 7/60 or x = 7/60 - 1/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) (131 (-1)^(1/3) + (30 sqrt(1462809) + 36253)^(2/3)) or x = 1/60 (131 (-1/(36253 + 30 sqrt(1462809)))^(1/3) + (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)) + 7/60